 Perfect discrimination of pure states requires orthogonality

If a positive operator-valued measure distinguishes normalized pure states $|u⟩,|v⟩$ with certainty, the effect $E$ for the first answer satisfies $⟨u|E|u⟩=1$ and $⟨v|E|v⟩=0$. Positivity of $E$ and $I-E$ implies $E|v⟩=0$ and $E|u⟩=|u⟩$, because a positive operator with zero expectation annihilates the vector. Consequently $⟨u|v⟩=⟨u|E|v⟩=0$. Conversely, orthogonal states can be measured with their rank-one linear projections. A common unitary operator preserves this inner product and cannot remove the obstruction.